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3 years ago

A car travels along a straight road to the east for

Physics

1 answer:

Korvikt [17]3 years ago

6 0

Answer: 90 m/s, 70 m/s

Explanation:

Given

A car travels 100 m east for 4 seconds and then turn around

goes back west for 50m in 1 sec

Distance traveled in the east is $100\times 4=400\ m$

Distance traveled in the west is $50\times 1=50\ m$

Total distance $=400+50=450\ m$

total displacement $=400-50=350\ m$

total time $t=4+1=5\ s$

Average speed

$\Rightarrow s_{avg}=\dfrac{\text{distance}}{\text{time}}=\dfrac{450}{5}\\\\\Rightarrow s_{avg}=90\ m/s$

Average velocity

$\Rightarrow v_{avg}=\dfrac{\text{displacement}}{\text{time}}=\dfrac{350}{5}\\\\\Rightarrow v_{avg}=70\ m/s$

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The electric field in a region of space increases from 0 to 2150 N/C in 5.00 s. What is the magnitude of the induced magnetic fi

Feliz [49]

To solve this problem we will use the Ampere-Maxwell law, which describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

$\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}$

Where,

B= Magnetic Field

l = length

$\mu_0$ = Vacuum permeability

$\epsilon_0$ = Vacuum permittivity

Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

$B(2\pi r) = \mu_0 \epsilon_0 \frac{d(EA)}{dt}$

Recall that the speed of light is equivalent to

$c^2 = \frac{1}{\mu_0 \epsilon_0}$

Then replacing,

$B(2\pi r) = \frac{1}{C^2} (\pi r^2) \frac{d(E)}{dt}$

$B = \frac{r}{2C^2} \frac{dE}{dt}$

Our values are given as

$dE = 2150N/C$

$dt = 5s$

$C = 3*10^8m/s$

$D = 0.440m \rightarrow r = 0.220m$

Replacing we have,

$B = \frac{r}{2C^2} \frac{dE}{dt}$

$B = \frac{0.220}{2(3*10^8)^2} \frac{2150}{5}$

$B =5.25*10^{-16}T$

Therefore the magnetic field around this circular area is $B =5.25*10^{-16}T$

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3 years ago

S To minimize neutron leakage from a reactor, the ratio of the surface area to the volume should be a minimum. For a given volum

erastovalidia [21]

To minimize neutron leakage from a reactor, the ratio of the surface area to the volume should be a minimum. For a given volume V the ratio of the sphere will be $\frac{4.83598}{c^{\frac{1}{3} } }$ .

We know that the surface area and volume of the sphere is given by:

$A=4 \pi r^{2}\\V=\frac{4}{3} \pi r^{3}$

Therefore, the ratio between the surface area and the volume for the sphere will be:

$\frac{A}{V}=\frac{4 \pi r^{2}\\}{\frac{4}{3} \pi r^{3}}=\frac{3}{r}$

Equating the volume to the constant c, we will find the value of $r$ .

$V=c=\frac{4}{3} \pi r^{3}\\r= (\frac{3c}{4\pi} )^{\frac{1}{3} }$

Substituting the value of r in the ration between surface area and volume, we get:

$\frac{A}{V}=\frac{3}{ (\frac{3c}{4\pi} )^{\frac{1}{3} }}$

Calculating the constants, we get:

$\frac{4.83598}{c^{\frac{1}{3} } }$

Hence, the ration between surface area and volume is $\frac{4.83598}{c^{\frac{1}{3} } }$

To learn more about surface area and volume of sphere, refer to:

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1 year ago

If a car travels 600m west in 25 seconds, what is its velocity?

Elena L [17]

$v = \frac{ \:the covered distance}{time} \\$

$v = \frac{600}{25} = \frac{6 \times 100}{25} = 6 \times 4 = 24 \: \\$

$v = 24 \: \frac{m}{s} \\$

_________________________________

If west means the west of the axis x the velocity equal :

$- 24 \: \frac{m}{s} \\$

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3 years ago

A hole is punched at A in a plastic sheet by applying a 660-N force P to end D of lever CD, which is rigidly attached to the sol

olasank [31]

Answer:

hello your question lacks some data and required diagram

G = 77 GPa, т all = 80 MPa

answer : required diameter = 252.65 * 10-^3 m

Explanation:

Given data :

force ( P ) = 660 -N force

displacement = 15 mm

G = 77 GPa

т all = 80 MPa

i) Determine the required diameter of shaft BC

considering the vertical displacement ( looking at handle DC from free body diagram )

D' = 0.3 sin∅ , where D = 0.015

hence ∅ = 2.8659°

calculate the torque acting at angle ∅ of CD on the shaft BC

Torque = 660 * 0.3 cos∅

= 660 * 0.3 * cos 2.8659 = 198 * -0.9622 = 190.5156 N

hello attached is the remaining part of the solution

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3 years ago

Consider a variety of colors of visible light (say 400 nm to 700 nm) falling onto a pair of slits.

babymother [125]

Answer:

Explanation:

The relationship between angle and wavelength for maxima and minima in Young's double slit experiment is given by

For constructive interference

$d\sin \theta =m\lambda$

For Destructive interference

$d\sin \theta =(m+\frac{1}{2})\lambda$

where $\lambda =wavelength$

$d=slit\ width$

m=order of maxima and minima

for second order maxima i.e. $m=2$

For smallest separation taking $\lambda =400 nm, \theta =90^{\circ}$

$d\sin 90=2\times 400\times 10^{-9}$

$d=0.8\times 10^{-6}$

$d=0.8\mu m$

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3 years ago